Sylow p-Subgroups of Group of Order 2p

Theorem

Let $p$ be an odd prime.

Let $G$ be a group of order $2 p$.

Then $G$ has exactly one Sylow $p$-subgroup.

This Sylow $p$-subgroup is normal.

Proof 1

Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.

From the Fourth Sylow Theorem:

$n_p \equiv 1 \pmod p$

From the Fifth Sylow Theorem:

$n_p \divides 2 p$

that is:

$n_p \in \set {1, 2, p, 2 p}$

But $p$ and $2 p$ are congruent to $0$ modulo $p$

So:

$n_p \notin \set {p, 2 p}$

Also we have that $p > 2$.

Hence:

$2 \not \equiv 1 \pmod p$

and so it must be that $n_p = 1$.

It follows from Sylow $p$-Subgroup is Unique iff Normal that this Sylow $p$-subgroup is normal.

$\blacksquare$

Proof 2

Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.

From the First Sylow Theorem, there exists at least $1$ Sylow $p$-subgroup of $G$.

Let $P$ be such a Sylow $p$-subgroup of $G$.

The index of $P$ is $2$.

From Subgroup of Index 2 is Normal, $P$ is normal in $G$.

From Sylow $p$-Subgroup is Unique iff Normal it follows that there is only $1$ Sylow $p$-subgroup of $G$.

$\blacksquare$

Proof 3

This is a specific instance of Group of Order $p q$ has Normal Sylow $p$-Subgroup, where $q = 2$.

$\blacksquare$